Isomorphism from $(\mathbb{Z}/200\mathbb{Z})^\times$

Question

I have to determine a direct sum of cyclic groups to which $(\mathbb{Z}/200\mathbb{Z})^\times$ is isomorphic. For general $n\in\mathbb{Z}_{\geq2}$, we can write $n=p_1^{e_1}\cdot\dots \cdot p_k^{e_k}$ for pairwise distinct prime numbers $p_i$ and positive exponents $e_i$. And from the Chinese remainder theorem we can conclude that $(\mathbb{Z}/n\mathbb{Z})^\times\cong(\mathbb{Z}/p_1^{e_1}\mathbb{Z})^\times\oplus\dotsb\oplus(\mathbb{Z}/p_k^{e_k}\mathbb{Z})^\times$. I learned that $(\mathbb{Z}/2^e\mathbb{Z})^\times$ is not cyclic for $e\geq3$, but $200=2^3\cdot5^2$, so that's where I get stuck.

Answer 1

If $p>2$, the group of units $\:(\mathbf Z/p^r\mathbf Z)^\times$ is cyclic.

On the other hand, the group of units $\:(\mathbf Z/2^s\mathbf Z)^\times$ is isomorphic to the direct product of the subgroup generated by the congruence class of of $-1$ (order $2$) and the subgroup generated by the class of $5$ (order $2^{s-2}$).

So here $$(\mathbf Z/200\mathbf Z)^\times\simeq(\mathbf Z/2^3\mathbf Z)^\times \times(\mathbf Z/5^2\mathbf Z)^\times\simeq(\mathbf Z/2\mathbf Z\times\mathbf Z/2\mathbf Z)\times\mathbf Z/20\mathbf Z.$$

Answer 2

You've stated that you already know that $$(\mathbb Z/200\mathbb Z)^*\cong (\mathbb Z/8\mathbb Z)^* \times (\mathbb Z/25\mathbb Z)^*,$$ so we just need to figure out how to write $(\mathbb Z/8\mathbb Z)^*$ and $(\mathbb Z/25\mathbb Z)^*$ as direct products of cyclic groups.

Let's start with $(\mathbb Z/8\mathbb Z)^*$. This is a group of order $\varphi(8)=4$ with elements $\{[1], [3], [5], [7]\}$. Since $$1^2\equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \pmod{8}$$ we see that every element of $(\mathbb Z/8\mathbb Z)^*$ has order dividing $2$. There is only one group of order $4$ with this property: $(\mathbb Z/2\mathbb Z) \times (\mathbb Z/2\mathbb Z)$. Thus $$(\mathbb Z/8\mathbb Z)^*\cong (\mathbb Z/2\mathbb Z) \times (\mathbb Z/2\mathbb Z).$$

Now let's consider $(\mathbb Z/25\mathbb Z)^*$. This is a group of order $\varphi(25)=20$. Consider the element $[2]$. You can easily check that the order of $[2]$ in $(\mathbb Z/25\mathbb Z)^*$ is $20$. This means that $(\mathbb Z/25\mathbb Z)^*$ must be cyclic of order $20$ generated by $[2]$. So we have $$(\mathbb Z/25\mathbb Z)^*\cong (\mathbb Z/20\mathbb Z) \cong (\mathbb Z/4\mathbb Z) \times (\mathbb Z/5\mathbb Z).$$

Putting all of this together we see that $$(\mathbb Z/200\mathbb Z)^* \cong (\mathbb Z/2\mathbb Z) \times (\mathbb Z/2\mathbb Z)\times (\mathbb Z/4\mathbb Z) \times (\mathbb Z/5\mathbb Z).$$